help-19

[south plume]

yea

[chilly marlin]

so 8

yes

[south plume]

next one

not sure what they mean by "the output of each gate symbolically"

but we can definitely do the first part

[chilly marlin]

okay :D

[south plume]

essentially it's just using this

[chilly marlin]

let me save this too

one sec

ok

[south plume]

so what happens to x and y in the beginning here

[chilly marlin]

it becomes x + y

[south plume]

yes

and what happens to z

[chilly marlin]

and the z becomes z compliment

[south plume]

yea

and so both of those go into AND

[chilly marlin]

so

(x+y)*zcompliment

[south plume]

yes

[south plume]

i think they mean writing x+y after the OR gate and z complement after the NOT gate

[chilly marlin]

yeah probably

[south plume]

so that's it

[chilly marlin]

oh wait

it says symbolically

maybe it means

write it with V or upside down V

[south plume]

i don't know about that

[chilly marlin]

oh yeah because those arent the symbols

[south plume]

this is more of a notation question than math question; it's something you decide with your teacher

[chilly marlin]

the symbols are + * compliment etc

[chilly marlin]

fair fair

what about the first part of the question?

is that just the end part

(x+y)*zcompliment

[south plume]

yeah

[chilly marlin]

oki :D

[south plume]

$(x+y)\cdot (\neg z)$

[clever fjord]

artemetra

[south plume]

or idk

this should be the answer

[chilly marlin]

yeah

[south plume]

as for the last one

sum-of-products means that there are no brackets left

so all you have are products that are being summed up, hence the name

[chilly marlin]

xz + yz

[south plume]

here use these laws

[chilly marlin]

uhm

[south plume]

yes

[chilly marlin]

this one?

[south plume]

no

[south plume]

actually idk this feels too easy lol

but it is correct

[chilly marlin]

LOL

ok :D

which law is that?

[chilly marlin]

so why are we using this law

over the other law?

[south plume]

cuz this is 2 variables

[south plume]

this is 3

they are not the same

[chilly marlin]

OHH

okok

got it

thank you so much

[south plume]

no problem

as you can see boolean algebra is really not that complicated

[chilly marlin]

yeah its pretty easy

only if my prof explained it

[south plume]

lol

[chilly marlin]

if you dont mind, can i add you?

[south plume]

yea go ahead

[chilly marlin]

okey thank you

[south plume]

if you are done type ".close"

[chilly marlin]

okey :)

thank you so much

bye bye

.close

[odd edge]

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[haughty gale]

bit confused here
$5x\sqrt{3}$

[clever fjord]

odokawa

[haughty gale]

derivate

my result was

$\frac{5x}{2\sqrt{3} + 5\sqrt{3}$

[clever fjord]

odokawa
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[warped glacier]

it's a constant * x

[haughty gale]

$\frac{5x}{2\sqrt{3}} + 5\sqrt{3}$

[zenith tartan]

5 root 3 is constant

[clever fjord]

odokawa

[haughty gale]

this was my result

but i was watching a video

and his answer was $5\sqrt{3}$

[clever fjord]

odokawa

[warped glacier]

yes that's correct

[haughty gale]

i've been studying derivatives for the last fucking 2 weeks and i dont fucking know how to do this shit

fuck

i dont know when tf i have to apply all those rules

my fuckin examen is tomorrow

fucking life

i see $5x\sqrt{3}$

[clever fjord]

odokawa

[haughty gale]

as product rule

this is so ugly damn shit

fucking maths of shit

.close

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[modest badge]

r is 3

[modest badge]

Small side is 10

Need to find R

Need help with this

[restive delta]

R is the hypotenuse is it?

[modest badge]

No radius of a circle

That is outside

And r is the radius that is inside

<@&286206848099549185>

[odd edge]

@modest badge Has your question been resolved?

[modest badge]

I didint draw the two circles

[modest badge]

Inscribed circles radius is 3

I nee to find the other one

<@&286206848099549185>

[wise idol]

(a+b-c)/2 is r

for a right angled triangle

use that

[modest badge]

I know r

r is 3

[wise idol]

put it in the formula

a is 10

[modest badge]

I need to find R

[wise idol]

id start with finding the sides

[modest badge]

I dont b or c

Tho

[wise idol]

a is 10, let b be x
c would be sqrt(100+x^2)

putting it in the formula would give a single variable equation

[modest badge]

Soo how do i write it

[wise idol]

(10+x-sqrt(100+x²))/2 = 3

[modest badge]

Why sqrt100 x

Tho

??

[wise idol]

when u put a,b,c in this

[wise idol]

thats c

using pythagoras

[modest badge]

Why is the sqrt tho

[wise idol]

sqrt(100+x²)

[wise idol]

x²+10²=c²

we need c, not c²

[modest badge]

X^2+100=100+x^2 ???

[wise idol]

ok we start from the beginning

you have a,b,c right

the 3 sides

[modest badge]

Yes

[wise idol]

c is the hypotenuse

[modest badge]

Yes

[wise idol]

we have a=10

we dont know b

[modest badge]

Yes

[wise idol]

and we dont know c either

if we apply pythagoras

a²+b²=c²

10²+b²=c²

100+b²=c²

we can solve for c in terms of b

c is sqrt(100+b²)

right

[modest badge]

Why the sqrt tho that’s what i dont get

Oo becous its not squared yet??

[wise idol]

this is c²

the side length is c

so we square root both sides

[modest badge]

Ok so next what

[wise idol]

then the formula for inradius (the radius of circle inside the triangle)

(a+b-c)/2 = r

you know r and a

and you know c in terms of b

substitute in the values

[modest badge]

10+b-sqrt(100+b^2)/2=r

?

[wise idol]

yea

you can easily solve for b here

[modest badge]

How do i get rid of the sqrt tho

[wise idol]

by squaring both sides

simplify it a little bit first

[modest badge]

After i do all of this then what

How do j find R from that

[wise idol]

another property of right angled triangles is that R is half the hypotenuse

[modest badge]

Is thath a rule?

[wise idol]

yea

[modest badge]

So ifter i find sides jus divide hypotnus by 2

[wise idol]

the mid point of the hypotenuse of a right angled triangle is the circumcentre

[wise idol]

yep

[modest badge]

Ok im not home right now i will try it in a hour and let you know

[wise idol]

sure

[odd edge]

@modest badge Has your question been resolved?

[odd edge]

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[small zinc]

h(x) = f(g(x))

f(x) = x/4 +3
g(x) = x+1

What is the function formula for h(x)

[small zinc]

I get that f(g(x))= f(x+1) but I dont know after that

[restive delta]

Yeah so

Wherever there's 'x' in f(x), substitute it with 'x+1'

[small zinc]

Wdym

(x+1)/4?

[restive delta]

Yup

[small zinc]

oh

[restive delta]

+3 too

[small zinc]

is that it

[mystic saffron]

h(x) = (x + 1)/4 + 3

[restive delta]

That is it

[small zinc]

Thats very anticlimactic

[restive delta]

Math is anti climatic.

[small zinc]

Thanks for the help

[restive delta]

Np

[small zinc]

.close

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[near wedge]

how can i get from this step to this one?

[twilit smelt]

idk calculus but i think i can help

what does (v*v_x)_x mean?

@near wedge

[near wedge]

i dont know

[twilit smelt]

bruh

[near wedge]

its v times a dervative of x and then its a dervative of that

[twilit smelt]

oh ok

what's v_xx then

second derivative of x?

[near wedge]

its the double dervagive of x yes

do you understand the step?

[twilit smelt]

honestly not quite

[near wedge]

okay

[twilit smelt]

oh wait

[near wedge]

can anyone help me understand this step please

[twilit smelt]

vx^2 + vy^2 is just v^2

[near wedge]

.close

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[twilit smelt]

did you solve tha

t

whatever

[odd edge]

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[hardy canopy]

given sinθ=1/2 and π/2<θ<π find the other 5 trigonometric functions

[hardy canopy]

help 😭 i have no idea how to do this

[crude carbon]

find value of θ where it gives 1/2 in sin function

[hardy canopy]

I'm trying to comprehend but I simply can't

[crude carbon]

at 30 degrees or at π/6 sin gives 1/2

now just find cosθ tanθ ....

[hardy canopy]

OH

Thank you

[cosmic iris]

simple take π=180 if you divide it by 6 you get 30

[hardy canopy]

What is the π/2 < θ < π used for?

[crude carbon]

sorry i did read it wrong

the angle is 5π/6 as its asking for 2nd quadrant

[hardy canopy]

Ohhhhhhh

I get it now

it's second quadrant because 90° < θ and θ < 180

π/6 is the reference angle so sin5π/6 = 1/2

tysm omg

REALLY 🙏😫

.close

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[fleet coral]

I am trying to solve this question using spherical coordinates

But I dont know how to get the limits for ϕ

[boreal crag]

What did you get when you converted the equations into spherical coordinates?

[fleet coral]

I have we have sphere of radius 4

and a cone

[boreal crag]

That isn't spherical coordinates

That's an English description

[fleet coral]

I dont know how to convert it

[boreal crag]

Replace x, y, and z with their representations in spherical coordinates

[fleet coral]

This is what I have now

x goes from 0 to 2?

[boreal crag]

What?

That isn't in spherical coordinates and it isn't one of the equations

Convert the equations into spherical coordinates

[fleet coral]

you mean of the equation of the sphere?

or the cone?

[boreal crag]

Both

Seems like you did the sphere already

[fleet coral]

what about the cone?

[boreal crag]

You haven't done that one

[fleet coral]

yeah but like how do I convert it into spherical ?

[boreal crag]

.

[fleet coral]

Like this?

I think I have to find where the sphere and the cone intersect

then use those coordinates

@boreal crag

[boreal crag]

no i have no idea what's going on here

your equation is z = sqrt(3)sqrt(x^2+y^2)

[fleet coral]

yeah

[boreal crag]

ok

[boreal crag]

so you have to simplify this now

[fleet coral]

oaky

I got

sqrt(2)sin(theta)cos(phi)

and where do i go from here?

[boreal crag]

That's not even an equation

[odd edge]

@fleet coral Has your question been resolved?

[fleet coral]

wait, let me try to figure it out

[odd edge]

@fleet coral Has your question been resolved?

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[slim ravine]

Sorry that my handwriting is messy, but can someone tell me what I'm doing wrong?

[plush verge]

isn't the answer just A = 45 degrees

all the other answers here don't work

[slim ravine]

Omg

I'm stupid

[plush verge]

lmao

[slim ravine]

Thank you so much

[plush verge]

no problem

[slim ravine]

.close

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[blissful jolt]

hey @orchid willow question for you about this

[gleaming hedge]

so

[blissful jolt]

#help-9 message

you said to draw x in a cartesian plane

(x - 3pi)

so in a cartesian plane

we actually go negative 3 pi is that right

and negative is moving to the right

but in a graph

why do we interpret (x - 3pi) as +3pi instead?

[orchid willow]

You dont

You go backwards, but you note that for certain multiples of pi going backwards o forward gives you the same result

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[gleaming wharf]

Not sure where to begin

[odd edge]

@gleaming wharf Has your question been resolved?

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@gleaming wharf Has your question been resolved?

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[gleaming wharf]

.reopen

[odd edge]

✅

[odd edge]

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[median cedar]

theyre similar right

by AA

RXC and ADC. triangles are similar?

[tame musk]

do you know the answer

somehow

that i would have been able to check my answer

@median cedar

[median cedar]

no

[tame musk]

oh

i have a idea

what if we draw another diagonal

from b ro d

so we have a triangle abm

[median cedar]

yeah

then what

[tame musk]

we can find bc

through pythagoras

we can find the area of abm

[plain frigate]

ABX and BCX is similar

if u diagonal a square then it will create two 45 degree equal slice angle

but cuz its a ractangle (one side is long, other side short) the diagnoal slice create one big slice angle and one littie slice angle

so u can see its corepsongin

[median cedar]

i still dk

how to do it

[plain frigate]

me niether lol

u can use sin cos tan maybe

[tame musk]

my idea went to pluto

let me try again

[median cedar]

how

oh wait

u can

[plain frigate]

also consider the sin cos tan of big triangle ACB

cuz it share the same angle

[tame musk]

tell me if the answer is 1.4

i did the math but i think it is wrong

guys we can get the area of abm

from abm we can get the another height

@median cedar Can i dm you

answer

the photo

[median cedar]

ok

[tame musk]

tell me if you got it

did you get it?

@median cedar

hello?

@median cedar

[median cedar]

thanks

[tame musk]

oh

[median cedar]

can you not see dm?

[tame musk]

oh

let me check

[median cedar]

i never thought of that

its correct i put your answer in and its correct

.cloes

.close

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[buoyant solstice]

im trying to prove that the orientation-preserving symmetry group for a tetrahedron is just the alternating group on 4 letters if we enumerate the vertices 1-4

[buoyant solstice]

i have identified all of the relevant transformations (consisting only of rotations) in my proof that makeup A_4, but now im wondering, do i need to prove something in regard to these being the only possible orientation-preserving transformations? and if so, how would i do that?

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@buoyant solstice Has your question been resolved?

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[pseudo oyster]

I have a question regarding matrix transposes since I'm learning SVDs right now. If A is an m x n matrix and x is an eigenvector for A^TA (A transpose times A) with the eigen value 0, how is it that A dotted with the vector x will also be a 0 vector?

[green elm]

so you have A^T A x = 0

multiply both sides by x^T

you get x^T A^T A x = 0

the left hand side is ||Ax||^2

which is zero, so the length of Ax is zero, so Ax is zero

[pseudo oyster]

Could you explain how the left has is ||Ax||^2 ?

||A||^2

[green elm]

sure

x^T A^T A x
is the same as
(Ax)^T Ax
which is the dot product of Ax with itself

which is the length squared of Ax

(assuming we're working with real matrices, not complex)

[pseudo oyster]

Got it

So this is always the case if the eigen value is 0 right?

[green elm]

yes

[pseudo oyster]

okay thank you so much

[green elm]

putting it another way, the null space of A^T A is the same as the null space of A

[pseudo oyster]

got it

.close

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[supple ore]

I also have one more after this

?????

[fair prism]

A factor for this case is of the form (x-a) so multiply that with x-2 and compare

[supple ore]

I dont understand

what should I do for my first step?

[fair prism]

Multiply (x-2)(x-a)

The goal is to find a

[supple ore]

is A the equation?

expression

[fair prism]

It is a number

[supple ore]

how do i find A?

[fair prism]

Expand it first

Then it may be clear but if not ill explain after

Oh shit I am blind as misread the expression

[supple ore]

?

[fair prism]

I thought the first term was x^2 not x^3 let me redo

[supple ore]

ok I dont really understand where to start still

[fair prism]

Ok so we are given that x-2 is a factor of x^3-bx^2-4

That means we can divide x-2 out of x^3-bx^2-4 and get another polynomial

This polynomial will be if degree 2, and we also know that the coefficient of x^2 will be 1, so overall it will be of the form x^2+cx+d

Is that all good with you?

[supple ore]

why would it be a degree of 2?

[wise idol]

use the fact that if (x-t) is a factor of f(x) then f(t)=0

[fair prism]

When you divide a polynomial of degree n with one of the degree m the new polynomial is degree n-m

Basically because x^3/x =x^2

[supple ore]

im fucking lost

how do you get a deree of 2 out of x^3+bx^2-4

[fair prism]

Because we can divide it by x-2

So I am saying that there exists a polynomial of degree 2, that when multiplied with (x-2), equals one with the form x^3-bx^2-4

Do you know polynomial long division or maybe synthetic division

[supple ore]

yes

i was about to do it then I got confused hpow I would do that with x^3+bx^2-4

[fair prism]

Ok if you want a more brute forcy way, divide x^3-bx^2-4 by (x-2)

[fair prism]

If u want to do it this way send what uve tried

[supple ore]

I would brind down 1

for the first correct?

im stuck here

[fair prism]

You would add b and 2 now

[supple ore]

I dont know what B is thouhj

[fair prism]

Yeah you will figure it out at the end of the entire process

For now just write b+2

And keep going

[fair prism]

You forgot about the +0x term

But u got the right idea

[odd edge]

@supple ore Has your question been resolved?

[odd edge]

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[arctic flare]

So Im doing a power series to find hte radiance of convergence and Ratio test fails since it = 1, The professor skipped only this one and its on a test review so my geuss is its some sort of edge case since we only ever have used R.T and

[sharp oak]

It's a p-series

Converges when the power ≥ 2

[arctic flare]

Thats what I figured. But How do I get the radiance from that?

Ohhhh Im dumb

I assume its by using the p > 1 constraint so 1 < 8x-3 <= inf,

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[storm crater]

I dont know where to start

[storm crater]

Since its in absolute value does that mean its just x^2+2x+4?

[glass hedge]

Plug in 0 for f(x) and solve for x

[storm crater]

but how do i deal with the -x^2

[fair prism]

not quite

[storm crater]

i know how to factor a quadratic but i forgot what to do if its a -x^2

[forest sky]

$|x| = \begin{cases} x, & , \text{if } x>0 \ -x & , \text{if } x<0 \end{cases}$

[clever fjord]

cloud

[fair prism]

remember |0|=0 so when |f(x)| is 0 so is f(x)

[storm crater]

so do i first do the top part

[fair prism]

and the converse is true too

[storm crater]

x, if x > 0

and then switch it

for the bottom piece

?

[forest sky]

we substitute in $-x^2+2x+4$ for $x$ here, so:

$\left| -x^2+2x+4 \right| = \begin{cases} \left(-x^2+2x+4\right), & \text{if } \left(-x^2+2x+4\right) >0 \ -\left(-x^2+2x+4\right), & \text{if } \left(-x^2+2x+4\right) <0 \end{cases}$

[clever fjord]

cloud

[storm crater]

but dont i still have to find the x intercepts

[forest sky]

yes. you can find where it switches between those two cases with the x-intercepts

[storm crater]

how do i factor -x^2+2x+4

[forest sky]

i don't think you can

[storm crater]

really??

[forest sky]

you can use the quadratic formula though

[storm crater]

it gives me an irrational number tho

[forest sky]

yes

[storm crater]

isnt that not allowed

to put that as the answer

-1.236067977

[forest sky]

I don't know how your homework is set up but if the answer is irrational then that's just what it is

[storm crater]

so if it just says that and it doesnt state whether to give the answer in a certain form i can just write the decimal i get from my calculator after using the formula?

[forest sky]

yes. then when you graph it, put it somewhere between the nearest gridlines

[storm crater]

ok ty

@forest sky this just a practize quiz example but if they gave a question like this whats stopping me from just putting it in my calculator then checking the right answer

[forest sky]

what do you mean by "putting it into your calculator"? mostly that depends on your class's policy on calculators

[storm crater]

like im allowed to use a ti-84 calculator at all times

even in exams

the graph function i mean

[forest sky]

you certainly could use your calculator for multiple-choice problems like this, but on longer-form exams you might have to graph it yourself and show your work on how you found that, in which case the calculator would be useful for checking your answer but not for showing your steps (which is usually required for credit)

i would recommend trying to work out the answers without calculator first, then come back and check your answers with calculator afterward

[storm crater]

ight

[forest sky]

in this question, for instance: there are only 2 graphs that even could involve an absolute value function out of the 4 possible answers

the only part you need to calculate is the x-intercept, which should be a quick calculation

so graphing it would actually be slower than working it out yourself

[topaz kestrel]

Guys, I need your help. Do any of you know any broker apps that can exchange money from one currency to another at a cheap rate, like using their government or bank rate?"

[storm crater]

tf

[storm crater]

wise.com 💯

.close

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[topaz kestrel]

Thanks but I need one made inside the USA

[odd edge]

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[civic isle]

help pls

i know that h,k of circle is 0, 0

and that its 2 right triangles

i think

[robust lagoon]

part (a) or part (b)?

[civic isle]

both 😦

or actually i can do b myself

idk how to start with a

[robust lagoon]

what do you need to get the equation of a line?

[civic isle]

do i do y-y1=m(x-x1)?

[robust lagoon]

yes but what is m?

[civic isle]

the derivative

ohh yeah i know where to start now ig

y^2=1-x^2

hol up

[robust lagoon]

have you done implicit differentiation before?

[civic isle]

yeah

[robust lagoon]

you'll need that to get dy/dx in terms of x and y

[civic isle]

dy/dx = 1-2x

or no

thats wrong

[robust lagoon]

nah don't think so

[civic isle]

2yy'= 1- 2x

[robust lagoon]

no 1

[civic isle]

ohhh

yeah

constant

y' = -x/y

[robust lagoon]

yeah nice

[civic isle]

is y1 and x1

0?

[robust lagoon]

well what's the point(s) on the curve you're considering?

[civic isle]

im lost

i know that radius is 1

but i dont think thats the point

[robust lagoon]

you want to find the equations of the tangent lines to the circle when x=a right

so we could find out what y can be when x=a

then that's the point in the equation of a line calculation

[civic isle]

apologies but i cant wrap my head around it

ahhh i have something i think

y= sqrt (1-a^2) ?

or y1

[robust lagoon]

yeah that's a solution

so one of the points of interest is (a, sqrt(1-a^2))

[civic isle]

y-(sqrt (1-a^2)) = -x/y (x-a)

[robust lagoon]

yeah so but -x/y is technically -x1/y1

the gradient depends on the point you're at here

[civic isle]

idk what a gradient is

[robust lagoon]

same thing as derivative

slope

[civic isle]

ah

so the equation is y-(sqrt (1-a^2)) = (-a/(1-a^2)) (x-a)

[robust lagoon]

yeah nice that's one

but there were two points right

if you think about the circle, for one x value we have 2 potential y-values

[civic isle]

yeah - and +

ahh

y-(sqrt (1-a^2)) = (-a/(1-a^2)) (x-a)
and
y+(sqrt (1-a^2)) = (-a/(1-a^2)) (x-a)

cuz the square root thingie

[robust lagoon]

yeah that's it

[civic isle]

would u mind helping me out with the 2nd part

HAHA

[robust lagoon]

wait actually have you subbed in the second y value in the derivative also

[civic isle]

oh

no i havent

ah wait yeah i wrote it wrong too

wait im lost

y-(+/-sqrt (1-a^2)) = (-a/(+/-sqrt(1-a^2)) (x-a)

[civic isle]

oh im not supposed to right?

cuz the derivative is -a/y

i got it now i think!

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[woven rampart]

Guys please help me with this one.

[tardy lagoon]

!status

[odd edge]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

@woven rampart Has your question been resolved?

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[scenic elm]

I know x1+x2 = a ; x1*x2 = b how do I calculate: x1^4+x2^4?

[restive delta]

,align x_1 + x_2 &=a \
x_1 x_2&=b

[clever fjord]

Lorentz

[scenic elm]

yes

[restive delta]

What have you tried

[tardy lagoon]

tip: it's similar to calculating x1^2+x2^2

[scenic elm]

makeing it from ((x1+x2)^2-2x1x2)^2

[restive delta]

Ye that's correct

[scenic elm]

that's the good answer?

[restive delta]

That would give you (x1)^2 + (x2)^2

[tardy lagoon]

no ((x1+x2)^2-2x1x2)^2 = x1^4+2x1^2x2^2+x2^4

[restive delta]

But you can use the same method to find ( x1)^4 + (x2)^4

[restive delta]

Ah, didn't see the last square

[scenic elm]

((x1+x2)^2-2x1x2)^2-2x1^2x2^2?

[tardy lagoon]

yep

[restive delta]

This is the term in the brackets

[scenic elm]

how could I make it from a,b?

[restive delta]

${({x_1}^2 + {x_2}^2)}^2$

[clever fjord]

Lorentz

[restive delta]

You can find this after finding x1 ^2 + x2 ^2

[scenic elm]

that task is to make it from a,b

without calculateing x1,x2

[restive delta]

Yeah

[tardy lagoon]

you just need to substitute x1+x2=a and x1x2=b to ((x1+x2)^2-2x1x2)^2-2x1^2x2^2

[scenic elm]

how to make this part? 2x1^2x2^2

[restive delta]

Well what's x1x2

[scenic elm]

b

[restive delta]

Yes

So (x1x2)^2 is?

[scenic elm]

2(b)^2

[restive delta]

Yeah

[scenic elm]

So I get (a^2-2b)^2-2(b)^2

right?

[restive delta]

One sec

Yeah thats right I think

[scenic elm]

So ((x1+x2)^2-2x1x2)^2-2x1^2x2^2=((a)^2-2b)^2-2b^2 where a = x1+x2 ; b = x1*x2

wait why is it like this in wa?

[restive delta]

${(a^2 - 2b)}^2 -2b^2$

[clever fjord]

Lorentz

[scenic elm]

Thanks for the help :)

[restive delta]

Np

[restive delta]

This is what you

Got right?

[scenic elm]

yes

[restive delta]

Alright

And it's .close

[scenic elm]

.close

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[oak arrow]

If a relation is not reflexive, does it automatically mean it is asymmetric ?

[weary pelican]

you need antisymmetry and irreflexiveness

[oak arrow]

Oh

[weary pelican]

irreflexive is stronger than "not reflexive", it means no element is in relation with itself

[waxen jackal]

Hi again

[oak arrow]

Hi pal

Thanks guys. Today is the day where I am going to solve the whoooole exercise.

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[odd edge]

@mystic saffron Has your question been resolved?

[odd edge]

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[gaunt palm]

Here is an assignment to put the math calculations from this week into practice. It will be due on Friday, December 8th at the end of the day (11:59pm). Refer to the calculation video which is uploaded in week 13 content area in order to complete the assignment. Use the information that is on the attached drawing, ignore the 5/8" plywood sheathing on the wall and treat the overhang dimension as 18" from outside of fascia to the framed wall, the same as I demonstrated in the video. The ridgeboard used in this drawing is 3 1/2" wide by 20" in depth (those are the exact dimensions), the same as it says on the drawing.

1. Find the birds mouth line length dimension

2. Find the total developed line length of the rafter.

3. Find Total rise above the wall plates with the given heel height on the drawing.

4. Find the dimension of the support post A under the ridge board.

5. What is the min and max allowable seat cut for this rafter?

https://cdn.discordapp.com/emojis/951166934043492372.gif?size=48&quality=lossless

i'm really bad in maths please help

[odd edge]

@gaunt palm Has your question been resolved?

[odd edge]

@gaunt palm Has your question been resolved?

[gaunt palm]

no

nooo

[odd edge]

@gaunt palm Has your question been resolved?

[odd edge]

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[merry beacon]

Why is it 4/2sqrt 13 or 2sqrt13/13

[honest turtle]

You multiply sqrt13 by numerator and denominator.

[merry beacon]

How ?

[honest turtle]

$$\frac{4}{2 \sqrt13}$$ $$=>\frac{4}{2 \sqrt13} \cdot \frac{1}{1}$$ $$=>\frac{4}{2 \sqrt13} \cdot \frac{\sqrt13}{\sqrt13}$$

[clever fjord]

Good

[merry beacon]

Thanks

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[restive rampart]

Are these steps correct?

I want to turn the function into a power series and then differentiate it

How do you convert the first function into a power series?

With the coefficent and the exponent?

I can do either by themselves but together idk what to do

wtf sqrt(3)??

I don't understand what I did wrong with differentiating the power series

??

[quiet berry]

the first sum should've been $\sum_{k=0}^{\infty}\left(-1\right)^{k}\left(3\right)^{k}\left(x^{2k}\right)$

[clever fjord]

Combustion

[restive rampart]

ohh

[quiet berry]

also you're differentiating incorrectly

[restive rampart]

How so?

[quiet berry]

the derivative should've been $\frac{-6x}{\left(1+3x^{2}\right)^{2}}$

[clever fjord]

Combustion

[quiet berry]

and here

(-1)^k doesn't have a variable in it

it's a constant coefficient

so is 3^k

[restive rampart]

I don't understand

What do I do then?

Do they change to 1?

[quiet berry]

what's the derivative of $ax$

[clever fjord]

Combustion

[restive rampart]

a

[quiet berry]

yep

a here is (-1)^k * 3^k

[restive rampart]

What do I do about the exponents then?

[quiet berry]

okay here's an example

$\frac{d}{dx}\sum_{n=0}^{\infty}a^{k}b^{k}c^{k}x^{k}=\sum_{n=0}^{\infty}a^{k}b^{k}c^{k}\left(kx^{\left(k-1\right)}\right)$

[clever fjord]

Combustion

[restive rampart]

They just get ignored?

[quiet berry]

unless they have an x, yes

[restive rampart]

Also, if I have x^(2k), does the 2 drop down with the k?

[quiet berry]

yep

[restive rampart]

It works!

What about for integration?

Same thing but with x^(k+1)/(k+1)?

[quiet berry]

yep

[restive rampart]

What if the power series had mutiple x's?

Replace x^k with kx^(k-1) or the integration one

And leave constants?

[quiet berry]

depends?

if you have multiple x's you can just combine their powers

then integrate/differentiate

[restive rampart]

I see

What about these

Where there is a k outside of the exponent

Leave them?

[quiet berry]

pretty much yeah

we're differentiating with respect to x not with respect to k

[restive rampart]

Okay, thank you

The fact it was inside the power series confused me

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[restive rampart]

How do I figure out whether or not endpoints are included in the interval of convergence in this case?

I know it's -sqrt(1/3) to sqrt(1/3) but idk how to tell the endpoints

It doesn't seem as simple as the previous times I've done it

[odd edge]

@restive rampart Has your question been resolved?

[warped glacier]

substitute the endpoints into the summation

[restive rampart]

I get that but then what

I don't know what test to use

[warped glacier]

wait is k = -sqrt(1/3)

like do you substitute that in for k?

[restive rampart]

I substitute -sqrt(1/3) into x

In the picture it says sqrt(x), it really means just x

That was from me putting sqrt(1/3)

[warped glacier]

oh it's just x, ahhhh

cause sqrt(-1/3) is undefined so I thought, ok ok

yeah, then you can simplify

$-\frac{1}{3} k (-3)^k (-1/3)^k (-1/3)^{k - 1}$

[clever fjord]

south

[warped glacier]

$= -\frac{1}{3}k (-1)^k (-1/3)^{k - 1}$

[clever fjord]

south

[warped glacier]

$= \frac{-1}{3}k (-1)^{2k - 1} (1/3)^{k - 1}$

[clever fjord]

south

[warped glacier]

by alternating series test, this converges

decreasing for all k > 3 (as |k/3| < 1) and terms tend to 0

[restive rampart]

What about the k not in the exponent

Wouldn't that make it go to infinity?

[warped glacier]

yeah, the key is k grows much much slower than 3^(k - 1)

rewrite that as k/3^(k - 1)

you would need another exponential function, like 3^k / 3^(k - 1) = 1/3 for it to diverge

[restive rampart]

I don't get that

wdym

I don't see how you would use lopitiles rule here

[warped glacier]

you can, cause it's an infinity/infinity form as k -> infinity

try LH then if you're not convinced

[restive rampart]

k/(3^k) basically?

[warped glacier]

yeah basically

[restive rampart]

Hmm okay I think I see

What test would you use for the positive endpoint then?

[warped glacier]

it's alternating series test again

[restive rampart]

Oh yeah duh

[warped glacier]

you can notice that cause the only difference between the positive and the negative

is that the power of (-1) is different

ye

[restive rampart]

So both converge then

[warped glacier]

so you basically know the answer for the positive endpoint

[restive rampart]

[-sqrt(1/3), sqrt(1/3)]

[warped glacier]

oh woops I should have put -sqrt(1/3)

doesn't matter too much

[warped glacier]

yes

[restive rampart]

Alright cool

Thank you

[warped glacier]

np!

[restive rampart]

Do you know why this is?

I don't see where the (k+1) came from

[warped glacier]

it's some indexing thing

wait what's f(x) again

what's the series representation?

[restive rampart]

This I think

[warped glacier]

right, so when you differentiate f(x)

you should get -6x / (1 + 3x^2)^2

differentiating that w.r.t x gives (-1)^k 3^k 2k x^(2k - 1)

[restive rampart]

Okay yeah I get that

[warped glacier]

hmmm

surely the k + 1 isn't correct

[restive rampart]

That's what it says is the correct answer

[warped glacier]

weird

oh right it's some series reindexing

cause the first term of f(x) is 1 right

that differentiates to 0

[restive rampart]

I don't understand what you mean

[warped glacier]

f(x) = 1/(1 + 3x^2) right

the first term of that series is 1

[restive rampart]

when x = 0

[warped glacier]

ok so if you put k = 1 into the bottom expression, you will get x^(-1)

that term doesn't exist cause the first term of f(x) differentiates to 0

so if you put k = 2, you get 9 * 2 * 2 * x

[warped glacier]

we need to divide this by -6 btw, cause of here

[restive rampart]

Yeah I see that. I don't see why that isn't in the answer though

[warped glacier]

if we divide this by -6, we will get $(-1)^{k - 1} 3^{k - 1} k x^{2k - 1}$

[clever fjord]

south

[warped glacier]

dividing everything by (-1) * 3 * 2

and then as k = 1 is undefined, we can replace k with k + 1 everywhere

so instead of starting from 2(0) - 1

we start from 2(1) - 1

yep that's exactly what happens in the answer

[restive rampart]

How did everything get ^(k-1)

[warped glacier]

this

divide everything by (-1) * 3 * 2

[restive rampart]

Why can't I just do what I tried in blue?

looks right, but different answer

[warped glacier]

cause if k = 0, you will have x^(-1)

that doesn't make sense for a power series

for example it's undefined at x = 0

[restive rampart]

oh

Then why isn't the other stuff ^(k+1)?

If you substitute k with (k+1)

Also where did the 2 go

I know there's a -6 being divided but I don't see how that relates to the 2 disappearing

Ohh

(-1)(3)(2)(-1)^k(3)^k(k+1)x^(2k+1)/(-6)

[warped glacier]

yes

[restive rampart]

Okay I see that makes sense now

I may have another question when it comes to differentiating/integrating power series

Do you mind if I ping you if that happens?

[warped glacier]

please find someone else

[restive rampart]

Alright

Thank you

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[thin void]

how did they get 34

[thin void]

i mean 36

for fv od

[odd magnet]

monthly for 3 years

3*12=36

[thin void]

oh shi

ik way wait

why is there a 12 there

for effective monthly rate

[odd magnet]

semi-anually compounded, once every 6 months or twice a year

1/6 or 2/12, same thing

[thin void]

okay ty

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[waxen owl]

Can someone show me how to plot this final graph in matlab?

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@waxen owl Has your question been resolved?

[waxen owl]

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[mortal hinge]

can someone help me with b ?

[mystic saffron]

make use of the power rule for logarithms

[
a\cd\m{\log_c}b = \m{\log_c}{b^a}
]

[mortal hinge]

is 2log_a(5) wrong ?

[lavish osprey]

sqrt is to the 1/2 power, not 2 🙂

[mortal hinge]

oh what

[mortal hinge]

i thought you had to undo the square root by squaring it 😭

[mystic saffron]

[
\s[m]{a^n} = a^{\ff nm}
]

[mortal hinge]

ohh

okay i got the answer, thank tou ☺️

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[minor bronze]

i would like the formula on how to do said question so i dont have to come back

[woven kelp]

There's not a formula as per say

We need to create an equation of some kind.

This can be done by finding the relationship between the angle 2x-6 and 4x+18

[minor bronze]

what im thinking is combine like terms, then go from there

[woven kelp]

This will help

[minor bronze]

yeah

its on the corresponding

[woven kelp]

You need to use 2 rules

1 isn't in that list

[minor bronze]

yeah

the textbook online given to me says to combine like terms then divide both sides by the new variable

[woven kelp]

Hmmm

[minor bronze]

so like

[woven kelp]

These two are the same

[minor bronze]

yes

[woven kelp]

So 2x-6+4x+18= ?

[minor bronze]

16?

you mean 6?

[woven kelp]

Angles on a straight line add to what?

[minor bronze]

180?

i have been shown that the answer will add up to 180

[woven kelp]

Yes

So 2x-6+4x+18=180

[minor bronze]

my issue is that i cant figure out the non-variables combined into one term

also, where are you getting 16?

[woven kelp]

Typo

[minor bronze]

ah

[woven kelp]

Send your working up to now as a picture pls

[minor bronze]

give me a sec

what i have is

6x - 24

i dont have a pic

not exactly sure what you mean

wait

X = 34?

thats what i got from multiplying 6 x 34 then subtracting 24

<@&286206848099549185>